Given two matrices $A, B$ of size $n \times n$ where $n \geqslant 2$
If $\det(A) =\det(B)$, can we infer that $\det(\operatorname{adj}(A)) = \det(\operatorname{adj}(B))$?
If so, how can we prove this?
My proof was intuitive, but I feel like it would not suffice.
$\det(\operatorname{adj}(A)) = \det(A)^{n-1}$ therefore… it is the same as $\det(B)^{n-1}$ and case closed.
But I think the whole purpose was to prove that $\det(\operatorname{adj}(A)) = \det(A)^{n-1}$ which I have no idea how to prove.
Best Answer
$$A.adj(A)=det(A)I_n$$
$$det(A.adj(A))=det(det(A)I_n)$$
$$det(A).det(adj(A))=det(A)^ndet(I_n)$$
$$det(A).det(adj(A))=det(A)^n.1$$
$$\Rightarrow det(adj(A))=\frac {det(A)^n} {det(A)}$$
$$\Rightarrow det(adj(A))= det(A)^{n-1}$$